CHAPTER 4 Differential and Integral operators The last part of the course includes some applications of Hilbert space and the spectral theorem { the completeness of the Fourier basis, some `Sturm-Liouville' theory, which is to say the spectral theory for second-order differential operators on an interval or the circle (this case is traditionally called Hill's equation) and enough of a treatment of the eigenfunctions for the harmonic oscillator to show that the Fourier transform is an isomorphism on L2(R): Once one has all this, one can do a lot more, but there is no time left. Such is life. 1. Fourier series Let us now try applying our knowledge of Hilbert space to a concrete Hilbert space such as L2(a; b) for a finite interval (a; b) ⊂ R: Any such interval with b > a can be mapped by a linear transformation onto (0; 2π) and so we work with this special interval. You showed that L2(a; b) is indeed a Hilbert space. One of the reasons for developing Hilbert space techniques originally was precisely the following result. Theorem 4.1. If u 2 L2(0; 2π) then the Fourier series of u; Z 1 X ikx −ikx (4.1) cke ; ck = u(x)e dx 2π (0;2π) k2Z converges in L2(0; 2π) to u: Notice that this does not say the series converges pointwise, or pointwise almost everywhere. In fact it is true that the Fourier series of a function in L2(0; 2π) converges almost everywhere to u; but it is hard to prove! In fact it is an important result of L. Carleson. Here we are just claiming that Z 1 X ikx 2 (4.2) lim ju(x) − cke j = 0 n!1 2π jk|≤n for any u 2 L2(0; 2π): Our abstract Hilbert space theory has put us quite close to proving this. First 0 2 observe that if ek(x) = exp(ikx) then these elements of L (0; 2π) satisfy Z Z 2π ( 0 0 0 if k 6= j (4.3) ekej = exp(i(k − j)x) = 0 2π if k = j: Thus the functions 0 ek 1 ikx (4.4) ek = 0 = p e kekk 2π 115 116 4. DIFFERENTIAL AND INTEGRAL OPERATORS form an orthonormal set in L2(0; 2π): It follows that (4.1) is just the Fourier-Bessel series for u with respect to this orthonormal set:- p 1 (4.5) c = 2π(u; e ) =) c eikx = (u; e )e : k k 2π k k k So, we already know that this series converges in L2(0; 2π) thanks to Bessel's in- equality. So `all' we need to show is 2 Proposition 4.1. The ek; k 2 Z; form an orthonormal basis of L (0; 2π); i.e. are complete: Z (4.6) ueikx = 0 8 k =) u = 0 in L2(0; 2π): This however, is not so trivial to prove. An equivalent statement is that the fi- 2 nite linear span of the ek is dense in L (0; 2π): I will prove this using Fej´er's method. In this approach, we check that any continuous function on [0; 2π] satisfying the additional condition that u(0) = u(2π) is the uniform limit on [0; 2π] of a sequence in the finite span of the ek: Since uniform convergence of continuous functions cer- tainly implies convergence in L2(0; 2π) and we already know that the continuous functions which vanish near 0 and 2π are dense in L2(0; 2π) this is enough to prove Proposition 4.1. However the proof is a serious piece of analysis, at least it seems so to me! There are other approaches, for instance we could use the Stone-Weierstrass Theorem; rather than do this we will deduce the Stone-Weierstrass Theorem from Proposition 4.1. Another good reason to proceed directly is that Fej´er's approach is clever and generalizes in various ways as we will see. So, the problem is to find the sequence in the span of the ek which converges to a given continuous function and the trick is to use the Fourier expansion that we want to check! The idea of Ces`arois close to one we have seen before, namely to make this Fourier expansion `converge faster', or maybe better. For the moment we can work with a general function u 2 L2(0; 2π) { or think of it as continuous if you prefer. The truncated Fourier series of u is a finite linear combination of the ek : 1 X Z (4.7) U (x) = ( u(t)e−iktdt)eikx n 2π jk|≤n (0;2π) where I have just inserted the definition of the ck's into the sum. Since this is a finite sum we can treat x as a parameter and use the linearity of the integral to write it as Z 1 X (4.8) U (x) = D (x − t)u(t);D (s) = eiks: n n n 2π (0;2π) jk|≤n Now this sum can be written as an explicit quotient, since, by telescoping, is=2 −is=2 i(n+ 1 )s −i(n+ 1 )s (4.9) 2πDn(s)(e − e ) = e 2 − e 2 : So in fact, at least where s 6= 0; i(n+ 1 )s −i(n+ 1 )s e 2 − e 2 (4.10) D (s) = n 2π(eis=2 − e−is=2) and the limit as s ! 0 exists just fine. 1. FOURIER SERIES 117 As I said, Ces`aro'sidea is to speed up the convergence by replacing Un by its average n 1 X (4.11) V (x) = U : n n + 1 l l=0 Again plugging in the definitions of the Ul's and using the linearity of the integral we see that n Z 1 X (4.12) V (x) = S (x − t)u(t);S (s) = D (s): n n n n + 1 l (0;2π) l=0 So again we want to compute a more useful form for Sn(s) { which is the Fej´er kernel. Since the denominators in (4.10) are all the same, n n is=2 −is=2 X i(l+ 1 )s X −i(l+ 1 )s (4.13) 2π(n + 1)(e − e )Sn(s) = e 2 − e 2 : l=0 l=0 Using the same trick again, n is=2 −is=2 X i(l+ 1 )s i(n+1)s (4.14) (e − e ) e 2 = e − 1 l=0 so is=2 −is=2 2 i(n+1)s −i(n+1)s 2π(n + 1)(e − e ) Sn(s) = e + e − 2 (4.15) 1 sin2( (n+1) s) =) S (s) = 2 : n n + 1 2 s 2π sin ( 2 ) Now, what can we say about this function? One thing we know immediately is that if we plug u = 1 into the discussion above, we get Un = 1 for n ≥ 0 and hence Vn = 1 as well. Thus in fact Z (4.16) Sn(x − ·) = 1; 8 x 2 (0; 2π): (0;2π) Looking directly at (4.15) the first thing to notice is that Sn(s) ≥ 0: Also, we can see that the denominator only vanishes when s = 0 or s = 2π in [0; 2π]: Thus if we stay away from there, say s 2 (δ; 2π − δ) for some δ > 0 then, sin(t) being a bounded function, −1 (4.17) jSn(s)j ≤ (n + 1) Cδ on (δ; 2π − δ): We are interested in how close Vn(x) is to the given u(x) in supremum norm, where now we will take u to be continuous. Because of (4.16) we can write Z (4.18) u(x) = Sn(x − t)u(x) (0;2π) where t denotes the variable of integration (and x is fixed in [0; 2π]): This `trick' means that the difference is Z (4.19) Vn(x) − u(x) = Sn(x − t)(u(t) − u(x)): (0;2π) 118 4. DIFFERENTIAL AND INTEGRAL OPERATORS For each x we split this integral into two parts, the set Γ(x) where x − t 2 [0; δ] or x − t 2 [2π − δ; 2π] and the remainder. So (4.20) Z Z jVn(x) − u(x)j ≤ Sn(x − t)ju(t) − u(x)j + Sn(x − t)ju(t) − u(x)j: Γ(x) (0;2π)nΓ(x) Now on Γ(x) either jt−xj ≤ δ { the points are close together { or t is close to 0 and x to 2π so 2π − x + t ≤ δ or conversely, x is close to 0 and t to 2π so 2π − t + x ≤ δ: In any case, by assuming that u(0) = u(2π) and using the uniform continuity of a continuous function on [0; 2π]; given > 0 we can choose δ so small that (4.21) ju(x) − u(t)j ≤ /2 on Γ(x): On the complement of Γ(x) we have (4.17) and since u is bounded we get the estimate Z −1 −1 (4.22) jVn(x) − u(x)j ≤ /2 Sn(x − t) + (n + 1) q(δ) ≤ /2 + (n + 1) q(δ) Γ(x) where q(δ) = 2 sin(δ=2)−2 sup juj is a positive constant depending on δ (and u): Here the fact that Sn is non-negative and has integral one has been used again to estimate the integral of Sn(x − t) over Γ(x) by 1: Having chosen δ to make the first term small, we can choose n large to make the second term small and it follows that (4.23) Vn(x) ! u(x) uniformly on [0; 2π] as n ! 1 under the assumption that u 2 C([0; 2π]) satisfies u(0) = u(2π): So this proves Proposition 4.1 subject to the density in L2(0; 2π) of the contin- uous functions which vanish near (but not of course in a fixed neighbourhood of) the ends.